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Katarina [22]
3 years ago
9

Mary starts at the edge of a circular platform that is slowly rotating on a frictionless axle. She then walks towards the opposi

te edge, passing through the platform center. Describe the motion of platform as Mary makes her trip. Explain the physics.
Physics
1 answer:
bulgar [2K]3 years ago
6 0

Answer:

Explanation:

The motion of Mary along the circular path is a centripetal.

As Mary moves from one edge of the circular platform to the other edge, she is covering a distance which is the radius of the circular path at a velocity.

According to the relationship

w = v/r where

w is the angular velocity

r is the radius

v is the linear velocity

Initially, before Mary starts, her linear speed is zero and her angular velocity is also zero. As she move towards the opposite edge, she is covering a distance of radius r. According to the formula, increase in radius will leads to decrease in her angular velocity and vice versa. As Mary starts moving towards the centre of the circular path, her angular velocity increases, at the centre of the platform, her angular velocity is at maximum at this point. As she moves further from the center to the other edge, her angular velocity decreases due to increase in distance covered across the circular path.

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Mazyrski [523]

Answer:

The diameter of the bull-wheel is 3.82

Explanation:

Given that,

Velocity = 2.0 m/s

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\omega=1.0472\ rad/s

We need to calculate the diameter of bull-wheel

Using formula of angular velocity

v= r\omega

r=\dfrac{v}{\omega}

Put the value into the formula

r=\dfrac{2.0}{1.0472}

r=1.91\ m

The diameter of the bull-wheel

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D=2\times1.91

D=3.82\ m

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3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

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1320 a -  420 a =14700

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a = 14700/900

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The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

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Answer:

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Explanation:

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Based upon the information you have learned throughout this module on blood spatter analysis, do you feel that analyzing blood s
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Answer:

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Explanation:

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